Seismic Wave Propagation: HW2 Due 13/5

Transcription

1 Reading: Scales Chapter , 6, 7, 9, Kallweit and Wood (198), Schneider (1978) Problems.1-.8 should be done on your own. You may work together to solve problems.9-.10, but you must hand in your own answers to get credit..1 Using the ray equation d d r=d n= show that rays are straight lines in homogeneous media. d dr = where d r ray trajectory, arclength, inde of refraction Since the material is homogenous =0 d d r d = d r, d r =k 1 r=k 1 k, the equation for a straight line. Show that in a medium where the inde of refraction only depends on depth that dr d =constant d r d = [d r d r yd y r ] d r d = [d r d y r y] without loss of generality a coordinate system may be chosen so that r =0 d r y is now the change in the horiontal coordinate of the ray relative to the change in arclength which is the same as sin i so d r d = sin i= c 0 c sin i =c p 0 by Snell's Law and where p is the ray parameter.3 The transport equation governs wave amplitude and is given by P P The Eikonal equation gives the solution for the phase. Use the Eikonal solution to calculate the solution for the amplitude assuming that we have a wave traveling in the + direction in homogeneous media. = k r=k P k P =0, k=k, k=0 P k=0 P=0 P=constant

2 .4 Make the substitution of k i for the 45 degree approimation in equation in Scales (in a similar manner as the 15 degree substitution in Scales eq ). Your answer should not have any derivatives in the denominator. [ i c ck i =[ c k c ck ] ] = [ c c ck ] k i c i ck = c k k i c i c = c 3 i = c 3c 4 c i 4.5 Derive the time domain version of the above equation by making the substitution i t i = c 3c 4 c i 4 i = 3 c 3 c 4 i c 4 i tt = i c 3c ttt i 4 c t i 4 tt = 1 c ttt 3c 4 t c 4

3 .6 Assume a stack of 3 horiontal layers. The first layer is 100 m thick and has a velocity of 1500 m/s, the second layer is 00 m thick and has a velocity of 500 m/s, the third layer is 300 m thick and has a velocity of 000 m/s. Below the third layer is a a half-space with velocity of 5000 m/s. Use the equation Z X p=p 0 d u p to calculate what take-off angle is required at the surface to generate refracted waves along the top of the half-space. Use the equation Z T = px u p d 0 to calculate the traveltime for the first refracted arrival from the half-space. Snell's Law: p= sin 1 v 1 = sin v = sin 3 v 3 = sin 4 v 4 Refraction will occur when sin 4 =, p=0.s/km Take-off angle 1 =sin 1 v 1 v 4 =sin 1 pv 1 =17.458degrees To calculate the traveltime, first calculate dx(p) for each layer, then calculate dt(p) for the each layer and sum the traveltimes. A MATLAB/Octave function that does the calculation for one layer at a time is given in layert_c.m. The total traveltime is then the sum of the traveltimes as outlined in the code below: layer=[0.1,1/1.5,1/.5; 0.,1/.5,1/.0; 0.3,1/.0,1/5.0] p=0., % ray parameter l=1; while l<=3 & irtr==1 % layers loop [d(l),dt(l),irtr(l)]=layert_c(p,layer(l,1),layer(l,),layer(l,3) ); =+*d(l); t=t+*dt(l); l=l+1; end % end of layers loop X=0.556 km T=0.65 s

4 .7 a) Show that, t Gt,={0 1 sin [ 0 t ]F 0, m 0 t} is the solution to the equation t ' ' 0 t =F t / m when F t={0 for t F 0 for t 0 for t } 0 The force acting on the system is nearly a delta function and is discontinuous. Therefore, the equation needs to be integrated to inspect if G is a solution: τ+ ϵ τ+ ϵ τ+ ϵ F 0 G ' ' dt +ω 0 G dt= τ ϵ τ ϵ τ ϵ m dt G' (τ +ϵ) G' (τ ϵ)+ω 0 ϵg(τ)= F ϵ 0 m G= F ϵ 0 ω 0 m sin [ω 0(t τ)], for t>τ ; =0, for t<τ G '= F ϵ 0 m cos[ω 0 (t τ)], for t> τ; =0, for t< τ cos[ω 0 (τ +ϵ)] cos[ω 0 (τ ϵ)]+ω 0 ϵsin [ω 0 τ]= =1, as ϵ 0 So G is a solution! b) What is the solution for (t) when F(t)=f(t) (t)= 1 mω 0 sin[ω 0 (t τ t ' )] f (t ')ϵdt '

5 .8 Write the corresponding Kirchhoff migration formula r,t=0= 1 cos '=0 Rc t ' s, y,0,t 'd ' dy' for D. Write an algorithm to do the migration in D. This algorithm will form the basis for your MATLAB/Octave migration code. You need to show a drawing indicating what the variables are that are in the algorithm. Assume that we can ignore the fact that the D response is different from the 3D response, then:,,t=0= 1 cos '=0 Rc t ' s ',0,t ' d' ' R, = 1 ' R c st ',R /c ' where st is the time derivative of the stack.9 Use MATLAB/Octave to generate a ero-offset section from the CDP gathers you produced in HW1 in process1.m. Do this by modifying the process.m script to include only data with offsets less 100 m in the stack. Plot the result and compare with the full stack. a) Why is the noise level higher in the ero-offset stack? The fold is lower b) Why does the ero-offset stack not etend as far as the full stack to the high CDP numbers The last source point was at CDP location 00 (4 km from the start of the section) so the farther offsets do not get included in the stack. c) Even if the noise level is higher in the ero-offset section, it shows less smearing in some of the reflections. Why? The ero-offset stack is not sensitive to velocity since NMO corrections are very small d) Use the command imagesc(,y,d) to get the correct scales on the vertical and horiontal aes. Label your plot using the label and ylabel commands. t=0:dt:; =0:0:0*99; imagesc(,t,stack_ero) e) Assume the background velocity to be 5000 m/s, What is the apparent dip of the reflection dipping towards lower CDP numbers? What is the true dip if the velocity is 5000 m/s? Apparent dip DT/DX = 1.13 s / 4000 m, DZ/DX= 875 m / 4000 m = 0.575: 35. degrees

6 True dip = asin(tan(apparent dip)): 45.4 degress

7 .10 Download the following to your computer: * MATLAB/Octave hw folder at Run the process3.m script. This script takes the stacked section and copies it into the array data_in and then migrates (phase shift migration) it using the velocity function v. The migration process takes several minutes depending upon what type of computer you are running on. Produce 3 different migrated sections using constant velocities of 3000, 5000 and 7000 m/s. Answer the following questions: a) Why are there dipping events towards the higher CDP numbers after migration? This is a wraparound effect from the Fourier transform in the -direction b) Why do the sections look different for the different migration velocities? The migration process is dependent upon velocity, the higher the velocity the more the reflections migrate c) Which migration velocity gives the best stack. Justify your answer. The one at 3000 m/s perhaps looks the best (migrates the data the least), but the one at 5000 m/s is more correct. The 7000 m/s migration produces significant artifacts.